Question

While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 29.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.

Answer 1

Complete Question

While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 29.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.

Calculate the distance between the stars.

Express your answer to two significant digits and include the appropriate units.

Answer:

The distance between the stars is $d =3.50*10^{30}m$

Explanation:

 From this question we are told that

         The orbital period is $29.3 \ days = 29.3 *24 *60* 60 = 210960\ sec$

         The mass of the stars are = Mass of sun $= 1.99*10^{30}kg$

         

For the two stars to keep on rotating on the circular orbit, the gravitational force must equal to the centripetal force and this can be mathematically represented as

                $\frac{Gm^2}{d^2} = \frac{mv^2}{r}$

Where r is the radius of the circular orbit

            G is the gravitational constant $=6.67 *10 ^{-11}$

           d  is the linear distance between the two stars = 2 r

This because during their oscillation around the circular orbit one usually hides the other

           $v = \frac{2 \pi r }{T}$

   Now substituting values into the above relation

              $\frac{Gm^2}{(2r)^2} = \frac{m[\frac{2 \pi r}{T} ]^2}{r}$

              $\frac{Gm^2}{(2r)^2} = \frac{m[\frac{4 \pi^2 r^2 }{T} ]}{r}$

                 $r^3 = [\frac{GmT^2}{16 \pi^2} ]$

Substituting values we have

                $r = [\frac{(6.67*10^{-11})(1.99*10^{30})(2531520)^2}{16(3.142)^2} ]^{\frac{1}{3} }$

                   $=\sqrt[3]{5.38529*10^{30}}$

                   $=1.75*10^{30}m$

Now d = 2r

=> d  $=2 *1.75*10^{30}m$

        $=3.50 *10^30m$